Two charges, Q1= 2.10 ?C, and Q2= 6.40 ?C are located at points (0,-2.50 cm ) an

Question

Two charges, Q1= 2.10 ?C, and Q2= 6.40 ?C are located at points (0,-2.50 cm ) and (0,+2.50 cm), as shown in the figure.

What is the magnitude of the electric field at point P, located at (5.50 cm, 0), due to Q1 alone?

What is the x-component of the total electric field at P?

What is the y-component of the total electric field at P?

What is the magnitude of the total electric field at P?

Now let Q2 = Q1 = 2.10 ?C. Note that the problem now has a symmetry that you should exploit in your solution. What is the magnitude of the total electric field at P?

Given the symmetric situation of the previous problem, what is the magnitude of the force on an electron placed at point P?

Explanation / Answer

E= (9*10^9 * Q)/ d^2

1. d= (5.52+2.52)

= 6.04cms = 0.0604m

so E= (9*109 * Q) / 0.06042

E= 5.192 * 1012 and its direction is 5.5x+ 2.5y

2. Now if more than 2 charges are there then we have to do vectorical addition

E(due to q2) = K * 6.4/0.06042 ( 5.5i- 2.5j )/6.04

E(due to q1) = K * 2.1/0.06042 (5.5i+ 2.5j )/6.04

So net Field is vectorical addition of above two so

X- component is (k / 0.06042 )* (2.1+6.4) (11 i/6.04) = 3.826 *1013

Y component is (k / 0.06042 )* (2.1- 6.4) (2.5 j/6.04)

In the symmetry case the Y component will be zero and the X component will be twice as the E of first question.

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